4.7 Applied optimization problems  (Page 5/8)

Step 1: Draw a rectangular box and introduce the variable $x$ to represent the length of each side of the square base; let $y$ represent the height of the box. Let $S$ denote the surface area of the open-top box.

Step 2: We need to minimize the surface area. Therefore, we need to minimize $S.$

Step 3: Since the box has an open top, we need only determine the area of the four vertical sides and the base. The area of each of the four vertical sides is $x·y.$ The area of the base is $x^2.$ Therefore, the surface area of the box is

Step 4: Since the volume of this box is $x^{2}y$ and the volume is given as $216\text{in}{.}^3,$ the constraint equation is

Solving the constraint equation for $y,$ we have $y=\frac{216}{x^2}.$ Therefore, we can write the surface area as a function of $x$ only:

Therefore, $S\left(x\right)=\frac{864}{x}+x^2.$

Step 5: Since we are requiring that $x^{2}y=216,$ we cannot have $x=0.$ Therefore, we need $x>0.$ On the other hand, $x$ is allowed to have any positive value. Note that as $x$ becomes large, the height of the box $y$ becomes correspondingly small so that $x^{2}y=216.$ Similarly, as $x$ becomes small, the height of the box becomes correspondingly large. We conclude that the domain is the open, unbounded interval $\left(0,\infty \right).$ Note that, unlike the previous examples, we cannot reduce our problem to looking for an absolute maximum or absolute minimum over a closed, bounded interval. However, in the next step, we discover why this function must have an absolute minimum over the interval $\left(0,\infty \right).$

Step 6: Note that as $x\to 0^{+},$ $S\left(x\right)\to \infty .$ Also, as $x\to \infty ,$ $S\left(x\right)\to \infty .$ Since $S$ is a continuous function that approaches infinity at the ends, it must have an absolute minimum at some $x\in \left(0,\infty \right).$ This minimum must occur at a critical point of $S.$ The derivative is

Therefore, $S^{\prime }\left(x\right)=0$ when $2x=\frac{864}{x^2}.$ Solving this equation for $x,$ we obtain $x^3=432,$ so $x=\sqrt[3]{432}=6\sqrt[3]{2}.$ Since this is the only critical point of $S,$ the absolute minimum must occur at $x=6\sqrt[3]{2}$ (see [link] ). When $x=6\sqrt[3]{2},$ $y=\frac{216}{{\left(6\sqrt[3]{2}\right)}^2}=3\sqrt[3]{2}\text{in}.$ Therefore, the dimensions of the box should be $x=6\sqrt[3]{2}\text{in}.$ and $y=3\sqrt[3]{2}\text{in}.$ With these dimensions, the surface area is